\displaystyle \textbf{MATHEMATICS (FOREIGN)}

\displaystyle \textbf{Series LRH/2}                                                                              \displaystyle \textbf{Code No. }30/2/1
\displaystyle \text{Roll No.}

  • \displaystyle \text{Candidates must write the Code on the title page of the answer-book.}
  • \displaystyle \text{Please check that this question paper contains 12 printed pages.}
  • \displaystyle \text{Code number given on the right hand side of the question paper should be}
  • \displaystyle \text{written on the title page of the answer-book by the candidate.}
  • \displaystyle \text{Please check that this question paper contains 30 questions.}
  • \displaystyle \text{Please write down the Serial Number of the question before}
  • \displaystyle \text{attempting it.}
  • \displaystyle \text{15 minutes time has been allotted to read this question paper. The question}
  • \displaystyle \text{paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the}
  • \displaystyle \text{student will read the question paper only and will not write any answer on}
  • \displaystyle \text{the answer script during this period.}

\displaystyle \text{Time allowed : 3 hours}                                                           \displaystyle \text{Maximum Marks : 80}

\displaystyle \textbf{General Instructions :}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) The question paper consists of 30 questions divided into four sections - A, B, C and D.}
\displaystyle \text{Section A comprises of ten questions of }1\text{ mark each, Section B comprises of five questions of}
\displaystyle 2\text{ marks each, Section C comprises of ten questions of }3\text{ marks each and Section D comprises}
\displaystyle \text{of five questions of }6\text{ marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact}
\displaystyle \text{requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, an internal choice has been provided in one question of}
\displaystyle 2\text{ marks each, three questions of }3\text{ marks each and two questions of }6\text{ marks each. You have to}
\displaystyle \text{attempt only one of the alternatives in all such questions.}
\displaystyle \text{(v) In question on construction, the drawings should be neat and exactly as per the given}
\displaystyle \text{measurement.}
\displaystyle \text{(vi) Use of calculators is not permitted.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{Questions number 1 to 10 carry 1 mark each.}

\displaystyle \textbf{Question 1. }\text{The HCF of 45 and 105 is 15. Write their LCM.}
\displaystyle \text{Answer:}
\displaystyle \text{HCF}\times\text{LCM}=\text{Product of two numbers}
\displaystyle 15\times\text{LCM}=45\times 105
\displaystyle \text{LCM}=\frac{45\times 105}{15}=315
\displaystyle \therefore\ \text{LCM}=315.
\\

\displaystyle \textbf{Question 2. }\text{If one zero of the polynomial }x^{2}-4x+1\text{ is }2+\sqrt{3}\text{, write the other}
\displaystyle \text{zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the coefficients of the polynomial are rational, irrational zeroes occur in conjugate pairs.}
\displaystyle \text{One zero}=2+\sqrt{3}
\displaystyle \therefore\ \text{Other zero}=2-\sqrt{3}.
\\

\displaystyle \textbf{Question 3. }\text{A tangent PQ at a point P of a circle of radius }5\text{ cm meets a line}
\displaystyle \text{through the centre O at a point Q so that OQ = }13\text{ cm. Find the}
\displaystyle \text{length PQ.}
\displaystyle \text{Answer:}
\displaystyle OP\perp PQ
\displaystyle \text{In right }\triangle OPQ,
\displaystyle OQ^{2}=OP^{2}+PQ^{2}
\displaystyle 13^{2}=5^{2}+PQ^{2}
\displaystyle PQ^{2}=169-25=144
\displaystyle PQ=12\text{ cm}
\displaystyle \therefore\ \text{Length of }PQ=12\text{ cm}.
\\

\displaystyle \textbf{Question 4. }\text{In Figure 1, MN }\parallel\text{ AB, BC = }7.5\text{ cm, AM = }4\text{ cm and MC = }2\text{ cm.}
\displaystyle \text{Find the length BN.} \displaystyle \text{Answer:}
\displaystyle MN\parallel AB
\displaystyle \therefore\ \triangle CMN\sim\triangle CBA
\displaystyle \frac{CN}{CB}=\frac{CM}{CA}
\displaystyle CA=AM+MC=4+2=6\text{ cm}
\displaystyle \frac{CN}{7.5}=\frac{2}{6}=\frac{1}{3}
\displaystyle CN=2.5\text{ cm}
\displaystyle BN=BC-CN=7.5-2.5=5\text{ cm}
\displaystyle \therefore\ BN=5\text{ cm}.
\\

\displaystyle \textbf{Question 5. }\text{If }6x=\sec\theta\text{ and }\frac{6}{x}=\tan\theta\text{, find the value of }9\left(x^{2}-\frac{1}{x^{2}}\right)\text{.}
\displaystyle \text{Answer:}
\displaystyle 6x=\sec\theta
\displaystyle \frac{6}{x}=\tan\theta
\displaystyle \sec^{2}\theta-\tan^{2}\theta=1
\displaystyle (6x)^{2}-\left(\frac{6}{x}\right)^{2}=1
\displaystyle 36x^{2}-\frac{36}{x^{2}}=1
\displaystyle 36\left(x^{2}-\frac{1}{x^{2}}\right)=1
\displaystyle 9\left(x^{2}-\frac{1}{x^{2}}\right)=\frac{1}{4}
\\

\displaystyle \textbf{Question 6. }\text{Find the distance between the points, A}(2a,6a)\text{ and B}(2a+\sqrt{3}a,5a)\text{.}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(2a+\sqrt{3}a-2a)^{2}+(5a-6a)^{2}}
\displaystyle =\sqrt{(\sqrt{3}a)^{2}+(-a)^{2}}
\displaystyle =\sqrt{3a^{2}+a^{2}}
\displaystyle =\sqrt{4a^{2}}
\displaystyle =2a
\displaystyle \therefore\ \text{Distance between the points}=2a.
\\

\displaystyle \textbf{Question 7. }\text{If the sum of first m terms of an A.P. is }2m^{2}+3m\text{, then what is its}
\displaystyle \text{second term ?}
\displaystyle \text{Answer:}
\displaystyle S_{m}=2m^{2}+3m
\displaystyle S_{1}=2(1)^{2}+3(1)=5
\displaystyle S_{2}=2(2)^{2}+3(2)=8+6=14
\displaystyle \text{Second term}=S_{2}-S_{1}=14-5=9
\displaystyle \therefore\ \text{Second term}=9.
\\

\displaystyle \textbf{Question 8. }\text{Find the value of k if P}(4,-2)\text{ is the mid point of the line segment}
\displaystyle \text{joining the points A}(5k,3)\text{ and B}(-k,-7)\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Using mid-point formula,}
\displaystyle P=\left(\frac{5k+(-k)}{2},\frac{3+(-7)}{2}\right)
\displaystyle (4,-2)=\left(\frac{4k}{2},\frac{-4}{2}\right)
\displaystyle (4,-2)=(2k,-2)
\displaystyle 2k=4
\displaystyle k=2
\\

\displaystyle \textbf{Question 9. }\text{The slant height of a frustum of a cone is }10\text{ cm. If the height of the}
\displaystyle \text{frustum is }8\text{ cm then find the difference of the radii of its two circular}
\displaystyle \text{ends.}
\displaystyle \text{Answer:}
\displaystyle l=10\text{ cm},\ h=8\text{ cm}
\displaystyle l^{2}=h^{2}+(R-r)^{2}
\displaystyle 10^{2}=8^{2}+(R-r)^{2}
\displaystyle (R-r)^{2}=100-64=36
\displaystyle R-r=6\text{ cm}
\displaystyle \therefore\ \text{Difference of the radii}=6\text{ cm}.
\\

\displaystyle \textbf{Question 10. }\text{A die is thrown twice. What is the probability that the same number}
\displaystyle \text{will come up either time ?}
\displaystyle \text{Answer:}
\displaystyle \text{Total possible outcomes}=6\times 6=36
\displaystyle \text{Favourable outcomes}=(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
\displaystyle \text{Number of favourable outcomes}=6
\displaystyle P(\text{same number on both throws})=\frac{6}{36}=\frac{1}{6}
\\

\displaystyle \textbf{SECTION B}
\displaystyle \text{Questions number 11 to 15 carry 2 marks each.}

\displaystyle \textbf{Question 11. }\text{If }-1\text{ and }2\text{ are two zeroes of the polynomial }
\displaystyle 2x^{3}-x^{2}-5x-2\text{, find }\text{its third zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the third zero be }\alpha.
\displaystyle \text{Sum of zeroes}=\frac{-(-1)}{2}=\frac{1}{2}
\displaystyle -1+2+\alpha=\frac{1}{2}
\displaystyle 1+\alpha=\frac{1}{2}
\displaystyle \alpha=-\frac{1}{2}
\displaystyle \therefore\ \text{Third zero}=-\frac{1}{2}.
\\

\displaystyle \textbf{Question 12. }\text{For what value of }k\text{ will the following pair of linear equations have no}
\displaystyle \text{solution ?}
\displaystyle 2x+3y=9;\qquad 6x+(k-2)y=(3k-2).
\displaystyle \text{Answer:}
\displaystyle 2x+3y-9=0
\displaystyle 6x+(k-2)y-(3k-2)=0
\displaystyle \text{For no solution,}
\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\ne\frac{c_{1}}{c_{2}}
\displaystyle \therefore\ \frac{2}{6}=\frac{3}{k-2}
\displaystyle \frac{1}{3}=\frac{3}{k-2}
\displaystyle k-2=9
\displaystyle k=11
\displaystyle \text{Now, }\frac{c_{1}}{c_{2}}=\frac{-9}{-(3k-2)}=\frac{9}{3k-2}
\displaystyle \text{For }k=11,\ \frac{9}{3k-2}=\frac{9}{31}\ne\frac{1}{3}
\displaystyle \therefore\ k=11.
\\

\displaystyle \textbf{Question 13. }\text{In an A.P., the first term is }-4\text{, the last term is }29\text{ and the sum of all}
\displaystyle \text{its terms is }150\text{. Find its common difference.}
\displaystyle \text{Answer:}
\displaystyle a=-4,\ l=29,\ S_{n}=150
\displaystyle S_{n}=\frac{n}{2}(a+l)
\displaystyle 150=\frac{n}{2}(-4+29)
\displaystyle 150=\frac{25n}{2}
\displaystyle n=12
\displaystyle l=a+(n-1)d
\displaystyle 29=-4+(12-1)d
\displaystyle 33=11d
\displaystyle d=3
\displaystyle \therefore\ \text{Common difference}=3.
\\

\displaystyle \textbf{Question 14. }\text{In Figure 2, a triangle ABC is drawn to circumscribe a circle of radius}
\displaystyle 3\text{ cm, such that the segments BD and DC into which BC is divided by}
\displaystyle \text{the point of contact D are of lengths }6\text{ cm and }8\text{ cm respectively. Find}
\displaystyle \text{the side AB if the area of }\triangle ABC=63\text{ cm}^{2}. \displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=rs
\displaystyle 63=3s
\displaystyle s=21\text{ cm}
\displaystyle BC=BD+DC=6+8=14\text{ cm}
\displaystyle s=\frac{AB+BC+AC}{2}
\displaystyle 21=\frac{AB+14+AC}{2}
\displaystyle AB+AC=28 \qquad (1)
\displaystyle \text{Tangents from an external point are equal.}
\displaystyle BF=BD=6\text{ cm and }CE=CD=8\text{ cm}
\displaystyle \text{Let }AF=AE=x.
\displaystyle AB=AF+BF=x+6
\displaystyle AC=AE+CE=x+8
\displaystyle \text{Using (1),}
\displaystyle x+6+x+8=28
\displaystyle 2x+14=28
\displaystyle x=7
\displaystyle AB=x+6=7+6=13\text{ cm}
\displaystyle \therefore\ AB=13\text{ cm}.
\\

\displaystyle \textbf{Question 15. }\text{Without using trigonometric tables, find the value of the following :}
\displaystyle \cot\theta\cdot\tan(90^\circ-\theta)-\sec(90^\circ-\theta)\mathrm{cosec}\theta+\sqrt{3}\cdot\tan12^\circ\cdot\tan60^\circ\cdot\tan78^\circ.

\displaystyle \textbf{OR}

\displaystyle \text{Find the value of }\sec45^\circ\text{ geometrically.}

\displaystyle \text{Answer:}
\displaystyle \cot\theta\cdot\tan(90^{\circ}-\theta)-\sec(90^{\circ}-\theta)\mathrm{cosec}\theta+\sqrt{3}\cdot\tan12^{\circ}\cdot\tan60^{\circ}\cdot\tan78^{\circ}
\displaystyle =\cot\theta\cdot\cot\theta-\mathrm{cosec}\theta\cdot\mathrm{cosec}\theta+\sqrt{3}\cdot\tan12^{\circ}\cdot\sqrt{3}\cdot\cot12^{\circ}
\displaystyle =\cot^{2}\theta-\mathrm{cosec}^{2}\theta+3
\displaystyle =-1+3
\displaystyle =2
\displaystyle \therefore\ \text{Required value}=2.

\displaystyle \textbf{OR}

\displaystyle \text{Consider a right isosceles }\triangle ABC\text{ with }\angle B=90^{\circ}\text{ and }AB=BC=a.
\displaystyle \text{Then }\angle A=\angle C=45^{\circ}.
\displaystyle AC=\sqrt{AB^{2}+BC^{2}}
\displaystyle AC=\sqrt{a^{2}+a^{2}}=a\sqrt{2}
\displaystyle \cos45^{\circ}=\frac{AB}{AC}=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}
\displaystyle \therefore\ \sec45^{\circ}=\frac{1}{\cos45^{\circ}}=\sqrt{2}
\displaystyle \therefore\ \sec45^{\circ}=\sqrt{2}.
\\

\displaystyle \textbf{SECTION C}
\displaystyle \text{Questions number 16 to 25 carry 3 marks each.}

\displaystyle \textbf{Question 16. }\text{Prove that }\sqrt{2}\text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sqrt{2}\text{ be a rational number.}
\displaystyle \therefore\ \sqrt{2}=\frac{a}{b},\text{ where }a\text{ and }b\text{ are coprime integers and }b\ne 0.
\displaystyle \text{Squaring both sides,}
\displaystyle 2=\frac{a^{2}}{b^{2}}
\displaystyle a^{2}=2b^{2}
\displaystyle \therefore\ a^{2}\text{ is even.}
\displaystyle \therefore\ a\text{ is even.}
\displaystyle \text{Let }a=2c.
\displaystyle (2c)^{2}=2b^{2}
\displaystyle 4c^{2}=2b^{2}
\displaystyle b^{2}=2c^{2}
\displaystyle \therefore\ b^{2}\text{ is even.}
\displaystyle \therefore\ b\text{ is even.}
\displaystyle \text{Thus, }a\text{ and }b\text{ have a common factor }2,\text{ which is a contradiction.}
\displaystyle \therefore\ \sqrt{2}\text{ is an irrational number.}
\\

\displaystyle \textbf{Question 17. }\text{Solve the following pair of linear equations for x and y :}
\displaystyle 2(ax-by)+(a+4b)=0;\qquad 2(bx+ay)+(b-4a)=0

\displaystyle \textbf{OR}

\displaystyle \text{A number consists of two digits. When the number is divided by the}
\displaystyle \text{sum of its digits, the quotient is }7\text{. If }27\text{ is subtracted from the}
\displaystyle \text{number, the digits interchange their places. Find the number.}

\displaystyle \text{Answer:}
\displaystyle 2ax-2by+a+4b=0 \qquad (1)
\displaystyle 2bx+2ay+b-4a=0 \qquad (2)
\displaystyle \text{Multiplying (1) by }a\text{ and (2) by }b,
\displaystyle 2a^{2}x-2ab y+a^{2}+4ab=0 \qquad (3)
\displaystyle 2b^{2}x+2ab y+b^{2}-4ab=0 \qquad (4)
\displaystyle \text{Adding (3) and (4),}
\displaystyle 2(a^{2}+b^{2})x+a^{2}+b^{2}=0
\displaystyle x=-\frac{1}{2}
\displaystyle \text{Multiplying (1) by }b\text{ and (2) by }a,
\displaystyle 2abx-2b^{2}y+ab+4b^{2}=0 \qquad (5)
\displaystyle 2abx+2a^{2}y+ab-4a^{2}=0 \qquad (6)
\displaystyle \text{Subtracting (5) from (6),}
\displaystyle 2(a^{2}+b^{2})y+4ab-4a^{2}-4b^{2}=0
\displaystyle 2(a^{2}+b^{2})y=4(a^{2}+b^{2})-4ab
\displaystyle y=2-\frac{2ab}{a^{2}+b^{2}}
\displaystyle \therefore\ x=-\frac{1}{2},\quad y=2-\frac{2ab}{a^{2}+b^{2}}.

\displaystyle \textbf{OR}

\displaystyle \text{Let the ten's digit be }x\text{ and unit's digit be }y.
\displaystyle \text{Number}=10x+y
\displaystyle \frac{10x+y}{x+y}=7
\displaystyle 10x+y=7x+7y
\displaystyle 3x-6y=0
\displaystyle x=2y \qquad (1)
\displaystyle 10x+y-27=10y+x
\displaystyle 9x-9y=27
\displaystyle x-y=3 \qquad (2)
\displaystyle \text{Using (1) in (2),}
\displaystyle 2y-y=3
\displaystyle y=3
\displaystyle x=6
\displaystyle \therefore\ \text{The number is }63.
\\

\displaystyle \textbf{Question 18. }\text{The sum of the first sixteen terms of an A.P. is }112\text{ and the sum of its}
\displaystyle \text{next fourteen terms is }518\text{. Find the A.P.}
\displaystyle \text{Answer:}
\displaystyle S_{16}=112
\displaystyle \frac{16}{2}[2a+(16-1)d]=112
\displaystyle 8(2a+15d)=112
\displaystyle 2a+15d=14 \qquad (1)
\displaystyle S_{30}=112+518=630
\displaystyle \frac{30}{2}[2a+(30-1)d]=630
\displaystyle 15(2a+29d)=630
\displaystyle 2a+29d=42 \qquad (2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle 14d=28
\displaystyle d=2
\displaystyle \text{Putting }d=2\text{ in (1),}
\displaystyle 2a+15(2)=14
\displaystyle 2a+30=14
\displaystyle 2a=-16
\displaystyle a=-8
\displaystyle \therefore\ \text{A.P. is }-8,-6,-4,-2,\ldots
\\

\displaystyle \textbf{Question 19. }\text{In }\triangle ABC\text{, right-angled at A, BL and CM are the two medians. Prove}
\displaystyle \text{that }4(BL^{2}+CM^{2})=5BC^{2}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(0,0),\ B(b,0)\text{ and }C(0,c).
\displaystyle \therefore\ BC^{2}=b^{2}+c^{2} \qquad (1)
\displaystyle \text{Since BL is a median, L is the mid-point of AC.}
\displaystyle L=\left(0,\frac{c}{2}\right)
\displaystyle BL^{2}=\left(b-0\right)^{2}+\left(0-\frac{c}{2}\right)^{2}
\displaystyle BL^{2}=b^{2}+\frac{c^{2}}{4} \qquad (2)
\displaystyle \text{Since CM is a median, M is the mid-point of AB.}
\displaystyle M=\left(\frac{b}{2},0\right)
\displaystyle CM^{2}=\left(0-\frac{b}{2}\right)^{2}+(c-0)^{2}
\displaystyle CM^{2}=\frac{b^{2}}{4}+c^{2} \qquad (3)
\displaystyle \text{Adding (2) and (3),}
\displaystyle BL^{2}+CM^{2}=b^{2}+\frac{c^{2}}{4}+\frac{b^{2}}{4}+c^{2}
\displaystyle BL^{2}+CM^{2}=\frac{5}{4}(b^{2}+c^{2})
\displaystyle 4(BL^{2}+CM^{2})=5(b^{2}+c^{2})
\displaystyle \therefore\ 4(BL^{2}+CM^{2})=5BC^{2}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 20. }\text{If }\tan\theta+\sin\theta=m\text{ and }\tan\theta-\sin\theta=n\text{, show that}
\displaystyle m^{2}-n^{2}=4\sqrt{mn}.

\displaystyle \textbf{OR}

\displaystyle \text{Show that :}
\displaystyle \left(1+\frac{1}{\tan^{2}\theta}\right)\left(1+\frac{1}{\cot^{2}\theta}\right)=\frac{1}{\sin^{2}\theta-\sin^{4}\theta}.

\displaystyle \text{Answer:}
\displaystyle m=\tan\theta+\sin\theta
\displaystyle n=\tan\theta-\sin\theta
\displaystyle m+n=2\tan\theta
\displaystyle m-n=2\sin\theta
\displaystyle m^{2}-n^{2}=(m+n)(m-n)
\displaystyle =2\tan\theta\cdot 2\sin\theta
\displaystyle =4\tan\theta\sin\theta
\displaystyle mn=(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)
\displaystyle =\tan^{2}\theta-\sin^{2}\theta
\displaystyle =\frac{\sin^{2}\theta}{\cos^{2}\theta}-\sin^{2}\theta
\displaystyle =\sin^{2}\theta\left(\frac{1-\cos^{2}\theta}{\cos^{2}\theta}\right)
\displaystyle =\sin^{2}\theta\cdot\frac{\sin^{2}\theta}{\cos^{2}\theta}
\displaystyle =\tan^{2}\theta\sin^{2}\theta
\displaystyle \sqrt{mn}=\tan\theta\sin\theta
\displaystyle \therefore\ m^{2}-n^{2}=4\sqrt{mn}.
\displaystyle \text{Hence proved.}

\displaystyle \textbf{OR}

\displaystyle \text{L.H.S.}=\left(1+\frac{1}{\tan^{2}\theta}\right)\left(1+\frac{1}{\cot^{2}\theta}\right)
\displaystyle =(1+\cot^{2}\theta)(1+\tan^{2}\theta)
\displaystyle =\mathrm{cosec}^{2}\theta\sec^{2}\theta
\displaystyle =\frac{1}{\sin^{2}\theta\cos^{2}\theta}
\displaystyle =\frac{1}{\sin^{2}\theta(1-\sin^{2}\theta)}
\displaystyle =\frac{1}{\sin^{2}\theta-\sin^{4}\theta}
\displaystyle =\text{R.H.S.}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 21. }\text{Draw a circle of radius }3\text{ cm. From a point P, }7\text{ cm away from the}
\displaystyle \text{centre of the circle, draw two tangents to the circle. Also, measure the}
\displaystyle \text{lengths of the tangents.}
\displaystyle \text{Answer:}
\displaystyle \text{Steps of construction:}
\displaystyle \text{1. Draw a circle with centre O and radius }3\text{ cm.}
\displaystyle \text{2. Mark a point P such that }OP=7\text{ cm.}
\displaystyle \text{3. Draw the perpendicular bisector of OP and let M be the mid-point of OP.}
\displaystyle \text{4. With M as centre and MO as radius, draw a circle cutting the given circle at A and B.}
\displaystyle \text{5. Join PA and PB. Then PA and PB are the required tangents.}
\displaystyle \text{Length of tangent}=\sqrt{OP^{2}-OA^{2}}
\displaystyle =\sqrt{7^{2}-3^{2}}=\sqrt{49-9}=\sqrt{40}=2\sqrt{10}\text{ cm}
\displaystyle \therefore\ PA=PB=2\sqrt{10}\text{ cm}\approx 6.3\text{ cm}.
\\

\displaystyle \textbf{Question 22. }\text{If point P}\left(\frac{1}{2},y\right)\text{ lies on the line segment joining the points A}(3,-5)
\displaystyle \text{and B}(-7,9)\text{, then find the ratio in which P divides AB. Also find the}
\displaystyle \text{value of y.}
\displaystyle \text{Answer:}
\displaystyle \text{Let P divide AB in the ratio }m:n.
\displaystyle \frac{-7m+3n}{m+n}=\frac{1}{2}
\displaystyle -14m+6n=m+n
\displaystyle 5n=15m
\displaystyle n=3m
\displaystyle \therefore\ m:n=1:3
\displaystyle y=\frac{9m-5n}{m+n}
\displaystyle y=\frac{9m-5(3m)}{m+3m}
\displaystyle y=\frac{-6m}{4m}=-\frac{3}{2}
\displaystyle \therefore\ \text{P divides AB in the ratio }1:3\text{ and }y=-\frac{3}{2}.
\\

\displaystyle \textbf{Question 23. }\text{Find the area of the shaded region in Figure 3, where a circular arc of}
\displaystyle \text{radius }7\text{ cm has been drawn with vertex O of an equilateral triangle}
\displaystyle \text{OAB, of side }12\text{ cm, as centre.}

\displaystyle \textbf{OR}

\displaystyle \text{The rain-water collected on the roof of a building, of}
\displaystyle \text{dimensions }22\text{ m}\times20\text{ m, is drained into a cylindrical vessel having}
\displaystyle \text{base diameter }2\text{ m and height }3.5\text{ m. If the vessel is full up to the}
\displaystyle \text{brim, find the height of rain-water on the roof. [Use }\pi=\frac{22}{7}\text{]}

\displaystyle \text{Answer:}
\displaystyle \angle AOB=60^{\circ}
\displaystyle \text{Area of shaded region}=\text{Area of circle}+\text{Area of equilateral triangle}-\text{Area of sector AOB}
\displaystyle =\pi(7)^{2}+\frac{\sqrt{3}}{4}(12)^{2}-\frac{60^{\circ}}{360^{\circ}}\times\pi(7)^{2}
\displaystyle =49\pi+36\sqrt{3}-\frac{49\pi}{6}
\displaystyle =\frac{245\pi}{6}+36\sqrt{3}
\displaystyle =\frac{245}{6}\times\frac{22}{7}+36\sqrt{3}
\displaystyle =128.33+62.35
\displaystyle =190.68\text{ cm}^{2}
\displaystyle \therefore\ \text{Area of shaded region}=190.68\text{ cm}^{2}.

\displaystyle \textbf{OR}

\displaystyle \text{Radius of cylindrical vessel}=1\text{ m}
\displaystyle \text{Volume of rain-water}=\pi r^{2}h
\displaystyle =\frac{22}{7}\times 1^{2}\times 3.5
\displaystyle =11\text{ m}^{3}
\displaystyle \text{Area of roof}=22\times 20=440\text{ m}^{2}
\displaystyle \text{Height of rain-water}=\frac{11}{440}=\frac{1}{40}\text{ m}
\displaystyle =0.025\text{ m}=2.5\text{ cm}
\displaystyle \therefore\ \text{Height of rain-water on the roof}=2.5\text{ cm}.
\\

\displaystyle \textbf{Question 24. }\text{Find the value of k for which the points A}(9,k)\text{, B}(4,-2)\text{ and C}(3,-3)
\displaystyle \text{are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{For collinear points, slopes AB and BC must be equal.}
\displaystyle \frac{-2-k}{4-9}=\frac{-3-(-2)}{3-4}
\displaystyle \frac{-2-k}{-5}=\frac{-1}{-1}
\displaystyle \frac{k+2}{5}=1
\displaystyle k+2=5
\displaystyle k=3
\displaystyle \therefore\ k=3.
\\

\displaystyle \textbf{Question 25. }\text{From a well-shuffled pack of playing cards, black jacks, black kings}
\displaystyle \text{and black aces are removed. A card is then drawn at random from the}
\displaystyle \text{pack. Find the probability of getting}
\displaystyle \text{(i) a red card,}
\displaystyle \text{(ii) not a diamond card.}
\displaystyle \text{Answer:}
\displaystyle \text{Total cards in a pack}=52
\displaystyle \text{Black jacks, black kings and black aces removed}=6
\displaystyle \text{Remaining cards}=52-6=46
\displaystyle \text{(i) Number of red cards}=26
\displaystyle P(\text{a red card})=\frac{26}{46}=\frac{13}{23}
\displaystyle \text{(ii) Number of diamond cards}=13
\displaystyle \text{Number of cards which are not diamonds}=46-13=33
\displaystyle P(\text{not a diamond card})=\frac{33}{46}
\\

\displaystyle \textbf{SECTION D}
\displaystyle \text{Questions number 26 to 30 carry 6 marks each.}

\displaystyle \textbf{Question 26. }\text{Some students planned a picnic. The total budget for food was}
\displaystyle \text{Rs. }2,000\text{. But }5\text{ students failed to attend the picnic and thus the cost}
\displaystyle \text{of food for each member increased by Rs. }20\text{. How many students}
\displaystyle \text{attended the picnic and how much did each student pay for the food ?}

\displaystyle \textbf{OR}

\displaystyle \text{Solve the following equation for }x\text{ :}
\displaystyle \frac{3x-4}{7}+\frac{7}{3x-4}=\frac{5}{2},\quad x\ne\frac{4}{3}.

\displaystyle \text{Answer:}
\displaystyle \text{Let the number of students originally planned be }x.
\displaystyle \text{Then the number of students who attended}=x-5.
\displaystyle \frac{2000}{x-5}-\frac{2000}{x}=20
\displaystyle \frac{2000x-2000(x-5)}{x(x-5)}=20
\displaystyle \frac{10000}{x(x-5)}=20
\displaystyle x(x-5)=500
\displaystyle x^{2}-5x-500=0
\displaystyle x^{2}-25x+20x-500=0
\displaystyle x(x-25)+20(x-25)=0
\displaystyle (x-25)(x+20)=0
\displaystyle x=25\text{ or }x=-20
\displaystyle \text{Since }x\text{ is positive, }x=25.
\displaystyle \text{Students attended}=25-5=20
\displaystyle \text{Amount paid by each student}=\frac{2000}{20}=100
\displaystyle \therefore\ \text{20 students attended the picnic and each paid Rs. }100.

\displaystyle \textbf{OR}

\displaystyle \frac{3x-4}{7}+\frac{5}{3x-4}=\frac{7}{2}
\displaystyle 2(3x-4)^{2}+70=49(3x-4)
\displaystyle 18x^{2}-48x+32+70=147x-196
\displaystyle 18x^{2}-195x+298=0
\displaystyle 18x^{2}-156x-39x+338=0
\displaystyle 6x(3x-26)-13(3x-26)=0
\displaystyle (3x-26)(6x-13)=0
\displaystyle 3x-26=0\text{ or }6x-13=0
\displaystyle x=\frac{26}{3}\text{ or }x=\frac{13}{6}
\displaystyle \therefore\ x=\frac{26}{3}\text{ or }\frac{13}{6}.
\\

\displaystyle \textbf{Question 27. }\text{If a line is drawn parallel to one side of a triangle to intersect the}
\displaystyle \text{other two sides in distinct points, prove that the other two sides are}
\displaystyle \text{divided in the same ratio.}
\displaystyle \text{Using the above, do the following :}
\displaystyle \text{In Figure 4, PQ }\parallel\text{ AB and AQ }\parallel\text{ CB. Prove that AR}^{2}=\text{PR}\cdot\text{CR.} \displaystyle \text{Answer:}
\displaystyle \text{Basic Proportionality Theorem:}
\displaystyle \text{If a line is drawn parallel to one side of a triangle to intersect the other two sides}
\displaystyle \text{in distinct points, then the other two sides are divided in the same ratio.}
\displaystyle \text{In }\triangle RPQ,\ AB\parallel PQ.
\displaystyle \therefore\ \frac{RA}{RP}=\frac{RB}{RQ} \qquad (1)
\displaystyle \text{In }\triangle RAQ,\ CB\parallel AQ.
\displaystyle \therefore\ \frac{RC}{RA}=\frac{RB}{RQ} \qquad (2)
\displaystyle \text{From (1) and (2),}
\displaystyle \frac{RA}{RP}=\frac{RC}{RA}
\displaystyle RA^{2}=RP\cdot RC
\displaystyle \therefore\ AR^{2}=PR\cdot CR
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 28. }\text{From a window }(9\text{ m above the ground})\text{ of a house in a street, the}
\displaystyle \text{angles of elevation and depression of the top and foot of another house}
\displaystyle \text{on the opposite side of the street are }30^\circ\text{ and }60^\circ\text{ respectively. Find}
\displaystyle \text{the height of the opposite house and the width of the street.}
\displaystyle \text{[Use }\sqrt{3}=1.732\text{]}

\displaystyle \textbf{OR}

\displaystyle \text{A vertical pedestal stands on the ground and is surmounted by a}
\displaystyle \text{vertical flag staff of height }5\text{ m. At a point on the ground the angles}
\displaystyle \text{of elevation of the bottom and the top of the flag staff are }30^\circ\text{ and}
\displaystyle 60^\circ\text{ respectively. Find the height of the pedestal.}

\displaystyle \text{Answer:}
\displaystyle \text{Let the width of the street be }x\text{ m and height of the opposite house be }h\text{ m.}
\displaystyle \tan60^{\circ}=\frac{9}{x}
\displaystyle \sqrt{3}=\frac{9}{x}
\displaystyle x=\frac{9}{\sqrt{3}}=3\sqrt{3}=3(1.732)=5.196\text{ m}
\displaystyle \tan30^{\circ}=\frac{h-9}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h-9}{3\sqrt{3}}
\displaystyle h-9=3
\displaystyle h=12\text{ m}
\displaystyle \therefore\ \text{Height of opposite house}=12\text{ m and width of street}=5.196\text{ m}.

\displaystyle \textbf{OR}

\displaystyle \text{Let the height of the pedestal be }h\text{ m and distance of the point from its foot be }x\text{ m.}
\displaystyle \tan30^{\circ}=\frac{h}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x}
\displaystyle x=h\sqrt{3} \qquad (1)
\displaystyle \tan60^{\circ}=\frac{h+5}{x}
\displaystyle \sqrt{3}=\frac{h+5}{x}
\displaystyle h+5=x\sqrt{3}
\displaystyle h+5=3h
\displaystyle 2h=5
\displaystyle h=\frac{5}{2}=2.5\text{ m}
\displaystyle \therefore\ \text{Height of the pedestal}=2.5\text{ m}.
\\

\displaystyle \textbf{Question 29. }\text{A container, open at the top, and made of a metal sheet, is in the form}
\displaystyle \text{of a frustum of a cone of height }24\text{ cm with radii of its lower and upper}
\displaystyle \text{ends as }7\text{ cm and }14\text{ cm respectively. Find the cost of milk which can}
\displaystyle \text{completely fill the container at the rate of Rs. }25\text{ per litre. Also find}
\displaystyle \text{the area of the metal sheet used to make the container. [Use }\pi=\frac{22}{7}\text{]}
\displaystyle \text{Answer:}
\displaystyle R=14\text{ cm},\ r=7\text{ cm},\ h=24\text{ cm}
\displaystyle \text{Volume of frustum}=\frac{1}{3}\pi h(R^{2}+r^{2}+Rr)
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times 24(14^{2}+7^{2}+14\times 7)
\displaystyle =8\times\frac{22}{7}(196+49+98)
\displaystyle =8\times\frac{22}{7}\times 343=8624\text{ cm}^{3}
\displaystyle =8.624\text{ litres}
\displaystyle \text{Cost of milk}=8.624\times 25=\text{Rs. }215.60
\displaystyle l=\sqrt{h^{2}+(R-r)^{2}}=\sqrt{24^{2}+7^{2}}=\sqrt{625}=25\text{ cm}
\displaystyle \text{Area of metal sheet}=\text{CSA of frustum}+\text{area of lower base}
\displaystyle =\pi(R+r)l+\pi r^{2}
\displaystyle =\frac{22}{7}(14+7)\times 25+\frac{22}{7}\times 7^{2}
\displaystyle =1650+154=1804\text{ cm}^{2}
\displaystyle \therefore\ \text{Cost of milk}=\text{Rs. }215.60\text{ and area of metal sheet}=1804\text{ cm}^{2}.
\\

\displaystyle \textbf{Question 30. }\text{If the mean of the following frequency distribution is }65.6\text{, find the}
\displaystyle \text{missing frequencies }(f_{1},f_{2})\text{ :}
\displaystyle \begin{array}{|c|c|} \hline \text{Class} & \text{Frequency} \\ \hline 10-30 & 5 \\ \hline 30-50 & 8 \\ \hline 50-70 & f_{1} \\ \hline 70-90 & 20 \\ \hline 90-110 & f_{2} \\ \hline 110-130 & 2 \\ \hline \text{Total} & 50 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle f_{1}+f_{2}+5+8+20+2=50
\displaystyle f_{1}+f_{2}=15 \qquad (1)
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class} & f & x & fx \\ \hline 10-30 & 5 & 20 & 100 \\ \hline 30-50 & 8 & 40 & 320 \\ \hline 50-70 & f_{1} & 60 & 60f_{1} \\ \hline 70-90 & 20 & 80 & 1600 \\ \hline 90-110 & f_{2} & 100 & 100f_{2} \\ \hline 110-130 & 2 & 120 & 240 \\ \hline \end{array}
\displaystyle \text{Mean}=\frac{\sum fx}{\sum f}
\displaystyle 65.6=\frac{2260+60f_{1}+100f_{2}}{50}
\displaystyle 3280=2260+60f_{1}+100f_{2}
\displaystyle 60f_{1}+100f_{2}=1020
\displaystyle 3f_{1}+5f_{2}=51 \qquad (2)
\displaystyle \text{Multiplying (1) by }3,
\displaystyle 3f_{1}+3f_{2}=45 \qquad (3)
\displaystyle \text{Subtracting (3) from (2),}
\displaystyle 2f_{2}=6
\displaystyle f_{2}=3
\displaystyle f_{1}=15-3=12
\displaystyle \therefore\ f_{1}=12,\quad f_{2}=3.
\\

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